How to set up the existence condition of an absolute value $$
\frac{\sqrt{4 + \arccos\left|\frac{2-x}{x+3}\right|}}{\sqrt{x^2 - 4x + 5} - 3}
$$
I'm trying to find the natural domain of the function above. I set up this conditions:
$$
\begin{cases}\sqrt{x^2 - 4x + 5} - 3\neq0&(denominator)\\x^2 - 4x + 5\ge0&(root)\\4 + \arccos\left|\frac{2-x}{x+3}\right|\ge0&(root)\\\left|\frac{2-x}{x+3}\right|\ge-1\cup\left|\frac{2-x}{x+3}\right|\le1&(arccos)\end{cases}
$$

Now I know that is necessary to set up two more conditions: the existence condition of the absolute value, and the existence condition of the fraction denominator. But I don't know how to deal with absolute values. Could someone explain it to me maybe using this example?
Note: These are not homework, but an exercise chose, because has an absolute value inside, to understand the theory.
 A: Given any real number $a$, $|a|$ always exists and is non-negative.
So all you need is $\frac{2-x}{x+3}$ to be well defined. Which translates to $x \neq -3$.
A: We first do it in a slow and tedious way, and then a quick way. For the $\arccos$ to be defined, we need $\frac{|2-x|}{|3+x|} \le 1$ (it is naturally non-negative).
Thus we need the inequality $|2-x|\le |x+3|$.  First suppose that $x \ge 2$. Then $|2-x|=x-2$, so we want $x-2 \le x+3$, which is true.
For $-3 <x<2$, we have $|2-x|=2-x$ and $|x+3|=x+3$. So we want $2-x\le x+3$, that is, $x \ge -1/2$.
At $x=-3$ things are clearly bad. But also if $x<-3$, then $|2-x|=2-x$ and $|x+3|=-(3+x)$. Thus our inequality becomes $2-x \le -(3+x)$, which is false. 
To sum up, the $\arccos$ part is defined if $x\ge -1/2$.  
Now we worry about the bottom. The function $x^2-4x+5$ is always positive, no trouble there. We want to make sure that $x^2-4x+5\ne 9$. We have equality at $-1$ and $5$, but $-1$ has already been ruled out.
Conclusion: Our function is defined for all $x\ge -1/2$ except $x=5$.
A quick way: The inequality $|2-x|\le |3+x|$ is equivalent to $(2-x)^2\le (3+x)^2$. Expand. The $x^2$'s cancel, and we get $x\ge -\frac{5}{10}$.
The tedious breaking up into cases of the first solution is unfortunately a useful tool in dealing with absolute value problems. The slick procedure of the "quick" solution is often much messier than doing things by cases.
